Lecture IV - Handling Data in more than one Dimension

Programming with Python

Author

Affiliation

Dr. Tobias Vlćek

Kühne Logistics University Hamburg - Fall 2026

Warm-up

Question 1

Three questions from Episode 3.

def announce(item):
    print(f"Fresh today: {item}")

result = announce("Mate")
print(result)

The second print shows what?

a) Fresh today: Mate b) Error c) None

Answer 1

def announce(item):
    print(f"Fresh today: {item}")

result = announce("Mate")
print(result)

Fresh today: Mate
None

c) None: announce has no return, so it hands back None. Printing something is not the same as returning it.

Question 2

def restock(n):
    n = n + 10
    return n

stock = 5
restock(stock)
print(stock)

a) 5 b) 15 c) Error

Answer 2

def restock(n):
    n = n + 10
    return n

stock = 5
restock(stock)
print(stock)

a) 5. n = n + 10 points the function’s own name n at a new number, so the global stock never moves. We never caught the result. To keep a result, assign it back.

Question 3

class Order:
    def __init__(self, item, qty, price):
        self.item = item
        self.qty = qty
        self.price = price

    def total(self):
        return self.qty * self.price

print(Order("Wrap", 2, 6.90).total())

a) 6.9 b) 13.8 c) 2

Answer 3

class Order:
    def __init__(self, item, qty, price):
        self.item = item
        self.qty = qty
        self.price = price

    def total(self):
        return self.qty * self.price

print(Order("Wrap", 2, 6.90).total())

13.8

b) 13.8. total() multiplies qty by price: 2 * 6.90, and print shows a float without the trailing zero. The object carries its own data, the method does the arithmetic.

Lists & Tuples

One name for many values

A list holds many values under one name, in order. That’s the cure for seventeen variables:

drinks = ["Mate", "Spezi", "Ayran"]
print(drinks)

['Mate', 'Spezi', 'Ayran']

Square brackets [], items separated by commas. Add a drink tomorrow and the list just grows. No new variable names.

Reaching in by position

Each item has an index. Python counts from 0:

drinks = ["Mate", "Spezi", "Ayran"]
print(drinks[0])   # the first item
print(drinks[2])   # the third item

Mate
Ayran

The first item is drinks[0], not drinks[1]. Off-by-one is a classic mistake of students.

Counting from the end

A negative index counts backwards from the end, no need to know the length:

drinks = ["Mate", "Spezi", "Ayran"]
print(drinks[-1])   # last item
print(drinks[-2])   # second to last

Ayran
Spezi

-1 is always the last item. Useful for getting “the most recent order”.

Slicing: a range of items

A slice list[start:stop] returns a new list, from start up to, but not including, stop:

drinks = ["Mate", "Spezi", "Ayran"]
print(drinks[0:2])   # items 0 and 1
print(drinks[-2:])   # the last two: negative start, open end

['Mate', 'Spezi']
['Spezi', 'Ayran']

The stop index is left out. drinks[0:2] gives you two items, not three. And leaving a side empty means “all the way”: drinks[-2:] starts two from the end and runs to the finish.

Predict: where does the slice stop?

A courier carries four cup sizes. What does this print?

sizes = ["S", "M", "L", "XL"]
print(sizes[1:3])

a) ['M', 'L', 'XL'] b) ['M', 'L'] c) ['S', 'M', 'L']

Predict. Pick a letter, then I reveal the answer.

Answer: the `stop` index is excluded

sizes = ["S", "M", "L", "XL"]
print(sizes[1:3])

['M', 'L']

b) ['M', 'L']. The slice starts at index 1 ("M") and stops before index 3, so index 3 ("XL") is never included. A slice from 1:3 gives you exactly 3 - 1 = 2 items.

Your turn

Open the exercise

Scan the QR or type the link:

python.tobiasvlcek.com/notebooks/ex_04_a/

QR code linking to the online notebook for exercise 4a, the same address as the link above.

Any questions? Let me know!

Growing a list

Lists are mutable: you can change them after creation. .append() adds one item to the end; + joins two lists into a new one:

drinks = ["Mate", "Spezi"]
drinks.append("Ayran")
print(drinks)

combined = ["Mate"] + ["Kombucha"]
print(combined)

['Mate', 'Spezi', 'Ayran']
['Mate', 'Kombucha']

.append() changes the list itself, and every name pointing at it sees that; + builds a fresh one. = never changes a list: it only points a name at one.

Predict: the second name

Tobi wants a winter copy of the drinks list to experiment on. What does the last line print?

summer = ["Mate", "Spezi"]
winter = summer
winter.append("Ayran")
print(summer)

a) ['Mate', 'Spezi'] b) ['Ayran', 'Mate', 'Spezi'] c) ['Mate', 'Spezi', 'Ayran']

Predict. Pick a letter, then I reveal the answer.

Answer: two names, one list

summer = ["Mate", "Spezi"]
winter = summer
winter.append("Ayran")
print(summer)

['Mate', 'Spezi', 'Ayran']

c) ['Mate', 'Spezi', 'Ayran']. winter = summer does not copy anything: both names point at the same list, so .append() through one name shows up under the other. A real copy is list(summer) or summer[:] (a slice with both ends open, so all of it).

How long is it?

len() tells you how many items a list holds:

drinks = ["Mate", "Spezi", "Ayran"]
print(len(drinks))

len() also works on strings (letters) and, soon, on dictionaries (keys).

Looping over positions

A loop over the list hands you each item. Sometimes you need to know where an item sits. Then loop over the positions instead:

drinks = ["Mate", "Spezi", "Ayran"]
for i in range(len(drinks)):
    print(i, drinks[i])

0 Mate
1 Spezi
2 Ayran

range(len(drinks)) counts 0, 1, 2: one number per position, and drinks[i] is the item that sits there.

Tuples: fixed-length records

A tuple looks like a list but uses () and cannot be changed, perfect for a record whose shape never varies, like opening hours (hour, minute):

opening = (9, 0)   # 9:00 sharp
print(opening[0])
print(opening[1])

hour, minute = opening   # unpack: one name per slot
print(hour, minute)

9
0
9 0

Indexing and slicing work exactly as on lists. You just can’t .append() to a tuple.

Your turn

Open the exercise

Scan the QR or type the link:

python.tobiasvlcek.com/notebooks/ex_04_b/

QR code linking to the online notebook for exercise 4b, the same address as the link above.

Any questions? Let me know!

Dictionaries & Sets

Look things up by name

A list finds items by position. But “what does a Spezi cost?” is a question about a name, not a position. A dictionary maps a key to a value:

prices = {"Mate": 3.50, "Spezi": 3.20, "Ayran": 2.80}
print(prices["Spezi"])

3.2

Curly braces {}, each pair written key: value. No more remembering that Spezi lives at index 1.

Building a dictionary

Keys are usually strings; values can be anything. Read a value with square brackets and its key:

prices = {"Mate": 3.50, "Spezi": 3.20, "Ayran": 2.80}
print(prices["Mate"])
print(prices["Ayran"])

3.5
2.8

The same square brackets as a list, but now the thing inside is a key, not a number.

The key that isn’t there

Ask for a key that doesn’t exist and Python raises a KeyError. It stops rather than guess:

prices = {"Mate": 3.50, "Spezi": 3.20}
print(prices["Cola"])
# KeyError: 'Cola'

A KeyError means the key is spelled wrong or was never added. However, there is a smarter way to ask for a key.

A safer read with `.get()`

.get() returns None for a missing key instead of raising, and you can supply a fallback:

prices = {"Mate": 3.50, "Spezi": 3.20}
print(prices.get("Cola"))       # missing → None
print(prices.get("Cola", 0))    # missing → your fallback

None
0

Use [] when the key must be there; use .get() when it might not.

Predict: is it on the menu?

in asks whether a value is in a list: "Mate" in drinks is True. On a dictionary, what does it look at?

prices = {"Mate": 3.50, "Spezi": 3.20}
print("Spezi" in prices, 3.20 in prices)

a) True False b) True True c) False True

Predict. Pick a letter, then I reveal the answer.

Answer: `in` checks the keys

prices = {"Mate": 3.50, "Spezi": 3.20}
print("Spezi" in prices, 3.20 in prices)
print(3.20 in prices.values())

True False
True

a) True False. Membership on a dictionary looks at the keys only. 3.20 is a value, so Python says it is not there. To search the values, ask 3.20 in prices.values().

Updating and adding

Assigning to a key updates it if it exists, or adds it if it doesn’t. Work on a copy (dict()) when the original must survive:

prices = {"Mate": 3.50, "Spezi": 3.20}
winter = dict(prices)        # a copy: keep the original safe
winter["Spezi"] = 3.40       # update an existing key
winter["Kombucha"] = 4.20    # add a brand-new key
print(winter)
print(prices)                # untouched (the summer menu comes back in April)

{'Mate': 3.5, 'Spezi': 3.4, 'Kombucha': 4.2}
{'Mate': 3.5, 'Spezi': 3.2}

One syntax, two jobs. Python decides by whether the key is already present.

Predict: the missing drink

A customer orders a Cola. It’s not on the menu. What does this line do?

prices = {"Mate": 3.50, "Spezi": 3.20}
print(prices["Cola"])

a) prints None b) prints "" c) raises KeyError d) adds "Cola"

Predict. Pick a letter, then I reveal the answer.

Answer: reading with `[]`

prices = {"Mate": 3.50, "Spezi": 3.20}
print(prices["Cola"])

KeyError: 'Cola'

c) raises KeyError: square-bracket reads never invent a value and never add one. Reach for .get() when a key might be missing.

Sets: only the unique ones

A set keeps each value once. Feed it duplicates and they collapse. Perfect for counting distinct things, like today’s regulars:

visitors = ["nina", "tom", "nina", "ada", "tom"]
regulars = set(visitors)
print(regulars)        # order is arbitrary: a set has none
print(len(regulars))   # how many different people

{'tom', 'ada', 'nina'}
3

Five visits, three people. A set answers “how many different?”.

Can be created from a list, or with curly braces:

regulars = {"nina", "tom", "ada"}
print(regulars)

{'tom', 'ada', 'nina'}

Your turn

Open the exercise

Scan the QR or type the link:

python.tobiasvlcek.com/notebooks/ex_04_c/

QR code linking to the online notebook for exercise 4c, the same address as the link above.

Any questions? Let me know!

Nested Data

A dictionary inside a dictionary

Values can themselves be dictionaries. Each delivery zone carries its own fee and estimated time:

zones = {
    "Altstadt": {"fee": 1.50, "eta": 12},
    "Hafen": {"fee": 2.50, "eta": 20},
}
print(zones["Hafen"])

{'fee': 2.5, 'eta': 20}

One key ("Hafen") points to a whole dictionary. This is how real data grows: structures inside structures.

Reading nested values

To reach an inner value, use two keys in a row, outer first, then inner:

zones = {
    "Altstadt": {"fee": 1.50, "eta": 12},
    "Hafen": {"fee": 2.50, "eta": 20},
}
print(zones["Hafen"]["fee"])
print(zones["Altstadt"]["eta"])

2.5
12

Read it left to right: “in zones, take Hafen, then its fee.”

Your turn

Open the exercise

Scan the QR or type the link:

python.tobiasvlcek.com/notebooks/ex_04_d/

QR code linking to the online notebook for exercise 4d, the same address as the link above.

Any questions? Let me know!

From a loop to one line

You already know the loop that builds a list. A list comprehension is that same loop, written on one line:

counts = [2, 1, 3]

# the loop you know
doubled = []
for c in counts:
    doubled.append(c * 2)
print(doubled)

# the one-liner
doubled = [c * 2 for c in counts]
print(doubled)

[4, 2, 6]
[4, 2, 6]

Read it as “c * 2 for each c in counts”. Same result, less typing.

`round` inside a comprehension

Any expression can go in front of the for, including round(). Loyalty pricing: 10% off, cleaned to two decimals:

drink_prices = [3.50, 3.20, 2.80]

print([p * 0.9 for p in drink_prices])            # raw floats, ugly
print([round(p * 0.9, 2) for p in drink_prices])  # rounded, clean

[3.15, 2.8800000000000003, 2.52]
[3.15, 2.88, 2.52]

The raw version leaks a 2.88000...3 float. Wrapping each price in round(_, 2) is the same thing we already used for money.

Looping over a dictionary

.items() hands you the menu as (key, value) pairs. Give the for two names and each pair is unpacked into them, like hour, minute = opening:

prices = {"Mate": 3.50, "Spezi": 3.20, "Ayran": 2.80}
for name, price in prices.items():
    print(name, price)

Mate 3.5
Spezi 3.2
Ayran 2.8

One name per slot, left to right. Two names, two slots.

Comprehensions build dictionaries

Swap the brackets for braces and give a key: value. Now you rebuild a whole menu in one line:

prices = {"Mate": 3.50, "Spezi": 3.20, "Ayran": 2.80}
loyalty = {name: round(price * 0.9, 2) for name, price in prices.items()}
print(loyalty)

{'Mate': 3.15, 'Spezi': 2.88, 'Ayran': 2.52}

The same name, price loop, now inside braces. The whole discounted menu, no loop body in sight.

Predict: the comprehension

What does this build?

nums = [1, 2, 3, 4]
print([n * n for n in nums])

a) [1, 4, 9, 16] b) [2, 4, 6, 8] c) [1, 2, 3, 4]

Predict. Pick a letter, then I reveal the answer.

Answer: square each item

nums = [1, 2, 3, 4]
print([n * n for n in nums])

[1, 4, 9, 16]

a) [1, 4, 9, 16]. The expression n * n runs once per item, so each number becomes its own square. 2 * n would double them; the bare n would just copy the list.

Predict: comprehension with filter

Some dishes sold nothing today. A comprehension can carry an if after the for. How long is the result?

counts = [2, 0, 3, 0, 1]
print([c * 2 for c in counts if c > 0])

a) [4, 0, 6, 0, 2] b) [4, 6, 2] c) [2, 3, 1]

Predict. Pick a letter, then I reveal the answer.

Answer: the `if` drops items

counts = [2, 0, 3, 0, 1]
print([c * 2 for c in counts if c > 0])

[4, 6, 2]

b) [4, 6, 2]. The if is a filter: items that fail it never reach the expression, so the result is shorter than the input. c * 2 still runs on the survivors, which rules out [2, 3, 1].

Your turn

Open the exercise

Scan the QR or type the link:

python.tobiasvlcek.com/notebooks/ex_04_e/

QR code linking to the online notebook for exercise 4e, the same address as the link above.

Any questions? Let me know!

To the Lab

After the break: the lab

Head to the lab notebook: Episode 4: The Menu Grows Up
You’ll rebuild the menu as a dictionary, count the regulars with a set, discount the card with a comprehension, and steer a lost courier through a nested campus map to the dorms
It runs entirely in your browser: no setup, just click and code

Download your .py before you leave. Closing the tab without downloading loses your work, and downloading is exactly how you hand in the checkpoints.

Wrap-up

Three things to remember

Lists hold many values in order: index from 0, count back with -1, slice with a stop that’s excluded
Dictionaries look up a value by its key; a missing key raises KeyError, so reach for .get() when it might not be there; sets keep each value once
Comprehensions fit a loop in one line: [expr for x in things] for lists, {k: v for ...} for dicts, round() welcome inside

Next time, Episode 5 starts with Checkpoint 2: 40 minutes, no AI, everything from Sessions III-IV. Then Tobi rewrites the checkout at 3 AM on four energy drinks. What could go wrong? We learn to read tracebacks and catch errors before the customer does.

Literature

Books to start with

Downey, A. B. (2024). Think Python: How to think like a computer scientist (Third edition). O’Reilly. Link to free online version
Elter, S. (2021). Schrödinger programmiert Python: Das etwas andere Fachbuch (1. Auflage). Rheinwerk Verlag.

Nothing new here, but these are still great books to start with!

For more, see the literature list of this course.

EP 4: Menu Grows Up

Seventeen variables and counting

Warm-up

Question 1

Answer 1

Question 2

Answer 2

Question 3

Answer 3

Lists & Tuples

One name for many values

Reaching in by position

Counting from the end

Slicing: a range of items

Predict: where does the slice stop?

Answer: the stop index is excluded

Your turn

Open the exercise

Growing a list

Predict: the second name

Answer: two names, one list

How long is it?

Looping over positions

Tuples: fixed-length records

Your turn

Open the exercise

Dictionaries & Sets

Look things up by name

Building a dictionary

The key that isn’t there

A safer read with .get()

Predict: is it on the menu?

Answer: in checks the keys

Updating and adding

Predict: the missing drink

Answer: reading with []

Sets: only the unique ones

Your turn

Open the exercise

Nested Data

A dictionary inside a dictionary

Reading nested values

Your turn

Open the exercise

From a loop to one line

round inside a comprehension

Looping over a dictionary

Comprehensions build dictionaries

Predict: the comprehension

Answer: square each item

Predict: comprehension with filter

Answer: the if drops items

Your turn

Open the exercise

To the Lab

After the break: the lab

Wrap-up

Three things to remember

Literature

Books to start with

Answer: the `stop` index is excluded

A safer read with `.get()`

Answer: `in` checks the keys

Answer: reading with `[]`

`round` inside a comprehension

Answer: the `if` drops items